What could be the reason why the call function does not function properly as a sorting mechanism?

Question

What could be the reason why the call function does not function properly as a sorting mechanism?

Currently immersed in javascript the good parts, I stumbled across an interesting example by the author:

['d','c','b','a'].sort(function(a,b) {
  return a.localeCompare(b);
});

The expected behavior was observed. Intrigued, I attempted to take it one step further with this code - the next logical progression:

['d','c','b','a'].sort(String.prototype.localeCompare.call);

However, this resulted in an error:

TypeError: object is not a function

This leaves me pondering the reason behind it... Any thoughts or insights on this?

javascript v8

Answer 1

Answer №1

apply should be bound to localeCompare:

['d','c','b','a'].sort(Function.prototype.apply.bind(String.prototype.localeCompare));

The issue you are facing is due to passing sort Function.prototype.apply. In the absence of a specified this, it defaults to the global object (window in browser settings). Consequently, when sort attempts to execute the provided function, it will use apply with this referencing the global object, which is typically not a function. Hence, binding apply ensures that this always points to localeCompare.

Answer 2

apply should be bound to localeCompare:

['d','c','b','a'].sort(Function.prototype.apply.bind(String.prototype.localeCompare));

The issue you are facing is due to passing sort Function.prototype.apply. In the absence of a specified this, it defaults to the global object (window in browser settings). Consequently, when sort attempts to execute the provided function, it will use apply with this referencing the global object, which is typically not a function. Hence, binding apply ensures that this always points to localeCompare.

Answer 3

Answer №2

The main issue causing the failure is that when

String.prototype.localeCompare.call

is used, it actually points to Function.prototype.call rather than String.prototype.localeCompare as expected.

Answer 4

The main issue causing the failure is that when

String.prototype.localeCompare.call

is used, it actually points to Function.prototype.call rather than String.prototype.localeCompare as expected.

Answer 5

Answer №3

When passing a reference to the call function, it is crucial to ensure a connection with the localeCompare method that you intend to utilize. Otherwise, the sort function will call upon the referenced function as

String.prototype.localeCompare.call

, but in the global context instead of within a specific function object.

To address this issue, as highlighted by @icktoofay, it is necessary to bind the call function to localeCompare. Below is a simplified demonstration illustrating how this can be achieved without utilizing the bind method:

function simpleBind(fn, _this) {
    return function() {
        return fn.apply(_this, arguments);
    }
}

['d','c','b','a'].sort(simpleBind(Function.prototype.call, 
                                  String.prototype.localeCompare));

Answer 6

When passing a reference to the call function, it is crucial to ensure a connection with the localeCompare method that you intend to utilize. Otherwise, the sort function will call upon the referenced function as

String.prototype.localeCompare.call

, but in the global context instead of within a specific function object.

To address this issue, as highlighted by @icktoofay, it is necessary to bind the call function to localeCompare. Below is a simplified demonstration illustrating how this can be achieved without utilizing the bind method:

function simpleBind(fn, _this) {
    return function() {
        return fn.apply(_this, arguments);
    }
}

['d','c','b','a'].sort(simpleBind(Function.prototype.call, 
                                  String.prototype.localeCompare));

Answer 7

Answer №4

Great news, I figured it out! Check out this solution:

['d','x','z','a'].sort(String.prototype.localeCompare.call.bind(String.prototype.localeCompare))

It's interesting how the sort function ends up invoking apply on call, so binding it to the string object is necessary for it to work properly within the scope. ^_^

Answer 8

Great news, I figured it out! Check out this solution:

['d','x','z','a'].sort(String.prototype.localeCompare.call.bind(String.prototype.localeCompare))

It's interesting how the sort function ends up invoking apply on call, so binding it to the string object is necessary for it to work properly within the scope. ^_^

What could be the reason why the call function does not function properly as a sorting mechanism?

Answer №1

Answer №2

Answer №3

Answer №4

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